course phy 232 I'm not sure how the test was sent they just took it. I guess it will be buy mail if you haven't received it yet. Can you just tell me my grade when evr you grade it? Thank you.
......!!!!!!!!...................................
16:52:20 query doppler shift experiment (experiment was to be read and viewed only) **** explain why the frequency of the sound observed when the buzzer moves toward you is greater than that of the stationary buzzer and why this frequency is greater than that observed when the buzzer is moving away from you
......!!!!!!!!...................................
RESPONSE --> When the buzzer is ringing it emits sound waves at a given frequency but as the buzzer moves towards you the sound waves are faster compared to you than normal due to the velocity of the buzzer added on. The frequency when the buzzer moves away is less because the velocity average is less because the buzzer source is emitting sound as it moves away and the wavelength stays the same so the frequency is less because lower velocity divided by constant wavelength means lower freq.
.................................................
......!!!!!!!!...................................
16:54:10 ** The 'pulses' emitted by an approaching source in a certain time interval are all received in a shorter time interval, since the last 'pulse' is emitted closer to the source than the first and therefore arrives sooner than if the source was still. So the frequency is higher. If the source is moving away then the last 'pulse' is emitted further from the source than if the source was still, hence arrives later, so the pulses are spread out over a longer time interval and the frequency is lower. GOOD EXPLANATION FROM STUDENT: Well, for the purposes of this explanation, I am going to explain the movement of the buzzer in one dimention which will be towards and away. The buzzer is actually moving in a circle which means it exists in three dimentions but is moving in two dimentions with relation to the listener. However, using trigonometry we can determine that at almost all times the buzzer is moving either towards or away from the listener so I will explain this in terms of one dimention. }When a buzzer is 'buzzing' it is emitting sound waves at a certain frequency. This frequency appears to change when the buzzer moves toward or away from the listener but the actual frequency never changes from the original frequency. By frequency we mean that a certain number of sound waves are emitted in a given time interval (usually x number of cycles in a second). So since each of the waves travel at the same velocity they will arrive at a certain vantage point at the same frequency that they are emitted. So If a 'listener' were at this given vantage point 'listening', then the listener would percieve the frequency to be what it actually is. Now, if the buzzer were moving toward the listener then the actual frequency being emitted by the buzzer would remain the same. However, the frequency percieved by the listener would be higher than the actual frequency. This is because, at rest or when the buzzer is not moving, all of the waves that are emitted are traveling at the same velocity and are emitted from the same location so they all travel the same distance. But, when the buzzer is moving toward the listener, the waves are still emitted at the same frequency, and the waves still travel at the same velocity, but the buzzer is moving toward the listener, so when a wave is emitted the buzzer closes the distance between it and the listener a little bit and there fore the next wave emitted travels less distance than the previous wave. So the end result is that each wave takes less time to reach the listener than the previously emitted wave. This means that more waves will reach the listener in a given time interval than when the buzzer was at rest even though the waves are still being emitted at the same rate. This is why the frequency is percieved to be higher when the buzzer is moving toward the listener. By the same token, if the same buzzer were moving away from the listener then the actual frequency of the waves emitted from the buzzer would be the same as if it were at rest, but the frequency percieved by the listener will be lower than the actual frequency. This is because, again at rest the actual frequency will be the percieved frequency. But when the buzzer is moving away from the listener, the actual frequency stays the same, the velocity of the waves stays the same, but because the buzzer moves away from the listener a little bit more each time it emits a wave, the distance that each wave must travel is a little bit more than the previously emitted wave. So therefore, less waves will pass by the listener in a given time interval than if the buzzer were not moving. This will result in a lower percieved frequency than the actual frequency. **
......!!!!!!!!...................................
RESPONSE -->
.................................................
......!!!!!!!!...................................
16:54:17 query General College Physics and Principles of Physics: what is a decibel?
......!!!!!!!!...................................
RESPONSE -->
.................................................
......!!!!!!!!...................................
17:00:29 ** dB = 10 log( I / I0 ), where I is the intensity of the sound in units of power per unit area and I0 is the 'hearing threshold' intensity. MORE EXTENSIVE EXPLANATION FROM STUDENT: Sound is possible because we exist in a medium of air. When a sound is emitted, a concussive force displaces the air around it and some amount energy is transferred into kinetic energy as air particles are smacked away from the force. These particles are now moving away from the initial force and collide into other air particles and send them moving and ultimately through a series of collisions the kinetic energy is traveling out in all directions and the air particles are what is carrying it. The behavior of this kinetic energy is to travel in waves. These waves each carry some amount of kinetic energy and the amount of energy that they carry is the intensity of the waves. Intensities of waves are given as a unit of power which is watts per square meter. Or since the waves travel in all directions they move in three dimentions and this unit measures how many watts of energy hits a square meter of the surface which is measuring the intensity. But we as humans don't percieve the intensities of sound as they really are. For example, a human ear would percieve sound B to be twice as loud as sound A when sound B is actually 10 times as loud as sound A. Or a sound that is ... 1.0 * 10^-10 W/m^2 is actually 10 times louder than a sound that is 1.0 * 10^-11 W/m^2 but the human ear would percieve it to only be twice as loud. The decibel is a unit of intensity for sound that measures the intensity in terms of how it is percieved to the human ear. Alexander Graham Bell invented the decibel. Bell originally invented the bel which is also a unit of intensity for waves. The decibel is one tenth of a bel and is more commonly used. The formula for determing the intensity in decibels is ... Intensity in decibles = the logarithm to the base 10 of the sound's intensity/ I base 0 I base 0 is the intensity of some reference level and is usually taken as the minimum intensity audible to an average person which is also called the 'threshold of hearing'. Since the threshold of hearing is in the denominator, if a sound is this low or lower the resulting intensity will be 0 decibles or inaudible. **
......!!!!!!!!...................................
RESPONSE -->
.................................................
......!!!!!!!!...................................
17:00:52 gen phy what is the difference between the node-antinode structure of the harmonics a standing wave in a string and in an organ pipe closed at one end
......!!!!!!!!...................................
RESPONSE -->
.................................................
......!!!!!!!!...................................
17:03:12 ** in a string there are nodes at both ends so the harmonics are described the the configurations NAN, NANAN, NANANAN, etc.. In a pipe closed at one end there is a node at one end and an antinode at the other so the possible configurations are NA, NANA, NANANA, etc.. displacement nodes are at both ends of the string, so the structure is N &&& N, where &&& is any sequence of nodes and antinodes that results in an alternating sequence. The possibilities are NAN, NANAN, NANANAN, ... , containing 1, 2, 3, 4, ..., half-wavelengths in the length of the string. Possible wavelengths are therefore 2 L, 1 L, 2/3 L, ..., where L is string length. For an open organ pipe the configuration must be N &&& A. Possibilities include NA, NANA, NANANA, NANANANA, ..., containing 1, 3, 5, 7, ..., quarter-wavelengths. Possible wavelengths are therefore 4 L, 4/3 L, 4/5 L, 4/7 L, ... **
......!!!!!!!!...................................
RESPONSE -->
.................................................
......!!!!!!!!...................................
17:03:15 **** gen phy what are beats?
......!!!!!!!!...................................
RESPONSE -->
.................................................
......!!!!!!!!...................................
17:03:33 ** Beats are what happens when the two sounds are close in frequency. Beats occur when the combined sound gets louder then quieter then louder etc. with a frequency equal to the differences of the frequencies of the two sounds. **
......!!!!!!!!...................................
RESPONSE -->
.................................................
......!!!!!!!!...................................
17:05:00 **** query univ phy 16.54 (20.32 10th edition) steel rod 1.5 m why hold only at middle to get fund? freq of fund? freq of 1st overtone and where held? **** why can the rod be held only at middle to get the fundamental?
......!!!!!!!!...................................
RESPONSE --> For a steel rod, the ends must be antinodes but to get the fundamental wavelength we want to have the longest wavelength possible so .5 wavelength = length of rod so the wavelength = 3m. so v=f*wavelength 5941m/s / 3 = f = 1.980kHz I'm not quite sure how to calculate the freq of 1st overtone.
.................................................
......!!!!!!!!...................................
17:10:37 STUDENT SOLUTION WITH INSTRUCTOR COMMENTS: Since the rod is a rigid body it can form the smallest frequency corresponding to the largest wavelength only when held at the center. The smallest frequency corresponding to the largest wavelength is known as the fundamental. (a) the fingers act as a sort of pivot and the bar performs a teeter - totter motion on either side of the pivot **note: the motion is more like the flapping of a bird's wings, though it obviously doesn't result from the same sort of muscular action but rather from an elastic response to a disturbance, and the motion attenuates over time**. With this manner of motion, the ends have the greatest amount of amplitude, thus making them the antinodes. (b)Holding the rod at any point other than the center changes the wavelngth of the first harmonic causing one end of the rod to have a smaller wave motion than the other. Since the fundamental frequency is in essence the smallest frequency corresponding to the largest wavelength, offsetting Lwould produce an unappropriatley sized wavelength. (c)Fundamental frequency of a steel rod: ( 5941 m /s) / [(2/1)(1.5m) = 1980.33Hz (d) v / (2/2) L = 3960.67Hz --achieved by holding the rod on one end. INSTRUCTOR COMMENTS: The ends of the rod are free so it can vibrate in any mode for which the ends are antinodes. The greatest possible wavelength is the one for which the ends are antinodes and the middle is a node (form ANA). Since any place the rod is held must be a node (your hand would quickly absorb the energy of the vibration if there was any wave-associated particle motion at that point) this ANA configuration is possible only if you hold the rod at the middle. Since a node-antinode distance corresponds to 1/4 wavelength the ANA configuration consists of 1/2 wavelength. So 1/2 wavelength is 1.5 m and the wavelength is 3 m. At propagation velocity 5941 m/s the 3 m wavelength implies a frequency of 5941 m/s / (3 m) = 1980 cycles/sec or 1980 Hz, approx.. The first overtone occurs with antinodes at the ends and node-antinode-node between, so the configuration is ANANA. This corresponds to 4 quarter-wavelengths, or a full wavelength. The resulting wavelength is 1.5 m and the frequency is about 3760 Hz. }THE FOLLOWING INFORMATION MIGHT OR MIGHT NOT BE RELATED TO THIS PROBLEM: STUDENT RESPONSES WITH INSTRUCTOR COMMENTS INSERTED To find the amplitudes at 40, 20 and 10 cm from the left end: The amplitudes are: at 40 cm 0 at 20 cm .004m at 10 cm .002828 m I obtained my results by using the information in the problem to write the equation of the standing wave. Since the cosine function is maximum at 0, I substituted t=0 into the equation and the value of x that I wanted to find the amplitude for. ** wavelength = 192 m/2 / (240 Hz) = .8 m. Amplitude at x is .4 cm sin(2 `pi x / .8 m) **the maximum transverse velocity and acceleration at each of these points are found from the equation of motion: `omega = 2 `pi rad * 240 cycles / sec = 480 `pi rad/s. .004 m * 480 `pi rad/s = 6 m/s approx and .004 m * ( 480 `pi rad/s)^2 = 9000 m/s^2, approx. .0028 m * 480 `pi rad/s = 4 m/s approx and .0028 m * ( 480 `pi rad/s)^2 = 6432 m/s^2, approx. **
......!!!!!!!!...................................
RESPONSE --> The first overtone is when ana goes to anana so the wavelength is decreased by 1/2 to 1.5m which means the freq doubles when held at the center to 3960Hz.
.................................................
U뼍ϰy|Ө assignment #024 024. `Query 13 Physics II 07-20-2008
......!!!!!!!!...................................
17:40:02 query experiment to be viewed and read but not performed: transverse and longitudinal waves in aluminum rod what is the evidence that the higher-pitched waves are longitudinal while the lower-pitched waves are transverse?
......!!!!!!!!...................................
RESPONSE --> When the rod is pointed towards the microphone the higher pitched waves are apparently louder so they must be longitudinal waves. SO the lower pitched waves must be transverse waves.
.................................................
......!!!!!!!!...................................
17:41:46 STUDENT RESPONSE: The logitudinal waves had a higher velocity. That doesn't provide evidence that the high-pitched wave was longitudinal, since we didn't directly measure the velocity of those waves. The higher-pitches waves were damped out much more rapidly by touching the very end of the rod, along its central axis, than by touching the rod at the end but on the side. The frequency with which pulses arrive at the ear determines the pitch. The amplitude of the wave affects its intensity, or energy per unit area. For a given pitch the energy falling per unit area is proportional to the square of the amplitude. Intensity is also proportional to the square of the frequency. **
......!!!!!!!!...................................
RESPONSE --> By touching the end of the rod more dampening occured so that means the higher pitched waves were probably longitudinal.
.................................................
......!!!!!!!!...................................
17:42:27 Query General College Physics and Principles of Physics 12.08: Compare the intensity of sound at 120 dB with that of a whisper at 20 dB.
......!!!!!!!!...................................
RESPONSE -->
.................................................
......!!!!!!!!...................................
17:44:43 The intensity at 120 dB is found by solving the equation dB = 10 log(I / I_threshold) for I. We get log(I / I_threshold) = dB / 10, so that I / I_threshold = 10^(120 / 10) = 12and I = I_threshold * 10^12. Since I_threshold = 10^-12 watts / m^2, we have for dB = 120: I = 10^-12 watts / m^2 * 10^12 = 1 watt / m^2. The same process tells us that for dB = 20 watts, I = I_threshold * 10^(20 / 10) = 10^-12 watts / m^2 * 10^2 = 10^-10 watts / m^2. Dividing 1 watt / m^2 by 10^-10 watts / m^2, we find that the 120 dB sound is 10^10 times as intense, or 10 billion times as intense. A more elegant solution uses the fact that dB_1 - dB_2 = 10 log(I_1 / I_threshold) - ( 10 log(I_2 / I_threshold) ) = 10 log(I_1 / I_threshold) - ( 10 log(I_2 / I_threshold) ) = 10 {log(I_1) - log( I_threshold) - [ ( log(I_2) - log(I_threshold) ]} = 10 { log(I_1) - log(I_2)} = 10 log(I_1 / I_2). So we have 120 - 20 = 100 = 10 log(I_1 / I_2) and log(I_1 / I_2) = 100 / 10 = 10 so that I_1 / I_2 = 10^10.
......!!!!!!!!...................................
RESPONSE -->
.................................................
......!!!!!!!!...................................
17:44:56 Query gen phy 12.30 length of open pipe, 262 Hz at 21 C? **** gen phy What is the length of the pipe?
......!!!!!!!!...................................
RESPONSE -->
.................................................
......!!!!!!!!...................................
17:49:00 GOOD STUDENT SOLUTION First we must determine the velocity of the sound waves given the air temperature. We do this using this formula v = (331 + 0.60 * Temp.) m/s So v = (331 + 0.60 * 21) m/s v = 343.6 m/s The wavelength of the sound is wavelength = v / f = 343.6 m/s / (262 Hz) = 0.33 meters. So 262 Hz = 343.6 m/s / 4 * Length Length = 0.33 meters f = v / (wavelength) 262 Hz = [343 m/s] / (wavelength) wavelength = 1.3 m. So the wavelength is 1.3 m. If it's an open pipe then there are antinodes at the ends and the wavelength is 2 times the length, so length of the the pipe is about 1.3 m / 2 = .64 m, approx.. Had the pipe been closed at one end then there would be a node and one end and an antinode at the other and the wavelength of the fundamental would have therefore been 4 times the length; the length of the pipe would then have been 1.3 m / 4 = .32 m. **
......!!!!!!!!...................................
RESPONSE -->
.................................................
......!!!!!!!!...................................
17:53:49 **** Univ phy 16.72 (10th edition 21.32): Crab nebula 1054 A.D.;, H gas, 4.568 * 10^14 Hz in lab, 4.586 from Crab streamers coming toward Earth. Velocity? Assuming const vel diameter? Ang diameter 5 arc minutes; how far is it?
......!!!!!!!!...................................
RESPONSE --> I found the equation Fr= sqrt(c-v/c+v)*fs but I'm not sure how to solve for v. I know that once v is found that multipling times the distance will give the diameter but I cannot find v. And I dont know what arc means.
.................................................
......!!!!!!!!...................................
17:55:05 ** Since fR = fS ( 1 - v/c) we have v = (fR / fS - 1) * c = 3 * 10^8 m/s * (4.586 * 10^14 Hz) / (4.568 * 10^14 Hz) = 1.182 * 10^6 m/s, approx. In the 949 years since the explosion the radius of the nebula would therefore be about 949 years * 365 days / year * 24 hours / day * 3600 seconds / hour * 1.182 * 10^6 m/s = 3.5 * 10^16 meters, the diameter about 7 * 10^16 meters. 5 minutes of arc is 5/60 degrees or 5/60 * pi/180 radians = 1.4 * 10^-3 radians. The diameter is equal to the product of the distance and this angle so the distance is distance = diameter / angle = 7 * 10^16 m / (1.4 * 10^-3) = 2.4 * 10^19 m. Dividing by the distance light travels in a year we get the distance in light years, about 6500 light years. CHECK AGAINST INSTRUCTOR SOLUTION: ** There are about 10^5 seconds in a day, about 3 * 10^7 seconds in a year and about 3 * 10^10 seconds in 1000 years. It's been about 1000 years. So those streamers have had time to move about 1.177 * 10^6 m/s * 3 * 10^10 sec = 3 * 10^16 meters. That would be the distance of the closest streamers from the center of the nebula. The other side of the nebula would be an equal distance on the other side of the center. So the diameter would be about 6 * 10^16 meters. A light year is about 300,000 km/sec * 3 * 10^7 sec/year = 9 * 10^12 km = 9 * 10^15 meters. So the nebula is about 3 * 10^16 meters / (9 * 10^15 m / light yr) = 3 light years in diameter, approx. 5 seconds of arc is 5/60 of a degree or 5 / (60 * 360) = 1 / 4300 of the circumference of a full circle, approx. If 1/4300 of the circumference is 6 * 10^16 meters then the circumference is about 4300 times this distance or about 2.6 * 10^20 meters. The circumference is 1 / (2 pi) times the radius. We're at the center of this circle since it is from here than the angular diameter is observed, so the distance is about 1 / (2 pi) * 2.6 * 10^20 meters = 4 * 10^19 meters. This is about 4 * 10^19 meters / (9 * 10^15 meters / light year) = 4400 light years distant. Check my arithmetic. **
......!!!!!!!!...................................
RESPONSE --> The velocity can be found by v = (fR / fS - 1) * c. Arc seconds are # seconds divided by 60 as degrees. A light year is how much distance light travels in a year which is about 9*10^15m / light year.
.................................................
......!!!!!!!!...................................
17:57:19 **** query univ phy 16.66 (21.26 10th edition). 200 mHz refl from fetal heart wall moving toward sound; refl sound mixed with transmitted sound, 85 beats / sec. Speed of sound 1500 m/s. What is the speed of the fetal heart at the instant the measurement is made?
......!!!!!!!!...................................
RESPONSE --> fl=200000085=2000000000+85
.................................................
......!!!!!!!!...................................
18:00:22 . ** 200 MHz is 200 * 10^6 Hz = 2 * 10^8 Hz or 200,000,000 Hz. The frequency of the wave reflected from the heart will be greater, according to the Doppler shift. The number of beats is equal to the difference in the frequencies of the two sounds. So the frequency of the reflected sound is 200,000,085 Hz. The frequency of the sound as experienced by the heart (which is in effect a moving 'listener') is fL = (1 + vL / v) * fs = (1 + vHeart / v) * 2.00 MHz, where v is 1500 m/s. This sound is then 'bounced back', with the heart now in the role of the source emitting sounds at frequency fs = (1 + vHeart / v) * 2.00 MHz, the 'old' fL. The 'new' fL is fL = v / (v - vs) * fs = v / (v - vHeart) * (1 + vHeart / v) * 2.00 MHz. This fL is the 200,000,085 Hz frequency. So we have 200,000,085 Hz = 1500 m/s / (v - vHeart) * (1 + vHeart / v) * 2.00 MHz and v / (v - vHeart) * (1 + vHeart / v) = 200,000,085 Hz / (200,000,000 Hz) = 1.000000475. A slight rearrangement gives us (v + vHeart) / (v - vHeart) = 1.000000475 so that v + vHeart = 1.000000475 v - 1.000000475 vHeart and 2.000000475 vHeart = .000000475 v, with solution vHeart = .000000475 v / (2.000000475), very close to vHeart = .000000475 v / 2 = .000000475 * 1500 m/s / 2 = .00032 m/s, about .3 millimeters / sec. **
......!!!!!!!!...................................
RESPONSE --> Do not solve for vl when sound travels to heart, only when it travels back so fL = v / (v - vs) * fs should be used instead.
.................................................
Ȫg옳֏ assignment #025 025. `Query 14 Physics II 07-20-2008
......!!!!!!!!...................................
18:18:31 Query Principles of Physics and General College Physics 12.40: Beat frequency at 262 and 277 Hz; beat frequency two octaves lower.
......!!!!!!!!...................................
RESPONSE -->
.................................................
......!!!!!!!!...................................
18:18:57 The beat frequency is the difference in the frequencies, in this case 277 Hz - 262 Hz = 15 Hz. One ocatave reduces frequency by half, so two octaves lower would give frequencies 1/4 as great. The difference in the frequencies would therefore also be 1/4 as great, resulting in a beat frequency of 1/4 * 15 Hz = 3.75 Hz.
......!!!!!!!!...................................
RESPONSE -->
.................................................
......!!!!!!!!...................................
18:19:19 Query gen phy problem 12.46 speakers 1.8 meters apart, listener three meters from one and 3.5 m from the other **** gen phy what is the lowest frequency that will permit destructive interference at the location of the listener?
......!!!!!!!!...................................
RESPONSE -->
.................................................
......!!!!!!!!...................................
18:22:18 STUDENT SOLUTION: To solve this problem, I first realize that for destructive interference to occur, the path difference is an odd multiple of half of the wavelength(ex. 1/2, 3/2, 5/2). I used Fig.12-17 in the text to help visualize this problem. Part (A) of the problem asks to calculate the lowest frequency at which destructive interference will occur at the point where two loudspeakers are 2.5m apart and one person stands 3.0m from one speaker and 3.5m from the other. The text states that 'destructive interference occurs at any point whose distance from one speaker is greater than its distance from the other speaker by exactly one-half wavelength.' The path difference in the problem is fixed, therefore the lowest frequency at which destructive interference will occur is directly related to the longest wavelength. To calculate the lowest frequency, I first have to calculate the longest wavelength using the equation 'dL ='lambda/2, where `dL is the path difference. 'lambda=2*'dL =2(3.5m-3.0m)=1m Now I can calculate the frequency using f=v/'lambda. The velocity is 343m/s which is the speed of sound. f=343m/s/1m=343 Hz. Thus, the lowest frequency at which destructive interference can occur is at 343Hz. Keeping in mind that destructive interference occurs if the distance equals an odd multiple of the wavelength, I can calculate (B) part of the problem. To determine the next wavelength, I use the equation 'dL=3'lambda/2 wavelength=2/3(3.5m-3.0m) =0.33m Now I calculate the next highest frequency using the equation f=v/wavelength. f^2=343m/s/0.33m=1030Hz. I finally calculate the next highest frequency. 'del L=5/2 'lambda wavelength=0.20m f^3=343m/s/0.2m=1715 Hz. INSTRUCTOR EXPLANATION: The listener is .5 meters further from one speaker than from the other. If this .5 meter difference results in a half- wavelength lag in the sound from the further speaker, the peaks from the first speaker will meet the troughs from the second. If a half-wavelength corresponds to .5 meters, then the wavelength must be 1 meter. The frequency of a sound with a 1-meter wavelength moving at 343 m/s will be 343 cycles/sec, or 343 Hz. The next two wavelengths that would result in destructive interference would have 1.5 and 2.5 wavelengths corresponding to the .5 m path difference. The wavelengths would therefore be .5 m / (1.5) = .33 m and .5 m / (2.5) = .2 m, with corresponding frequencies 343 m/s / (.33 m) = 1030 Hz and 343 m/s / (.2 m) = 1720 Hz, approx. ****
......!!!!!!!!...................................
RESPONSE -->
.................................................
......!!!!!!!!...................................
18:22:30 **** gen phy why is there no highest frequency that will permit destructive interference?
......!!!!!!!!...................................
RESPONSE -->
.................................................
......!!!!!!!!...................................
18:22:36 ** You can get any number of half-wavelengths into that .5 meter path difference. **
......!!!!!!!!...................................
RESPONSE -->
.................................................
......!!!!!!!!...................................
18:23:01 gen phy what must happen in order for the sounds from the two speakers to interfere destructively, assuming that the sources are in phase?
......!!!!!!!!...................................
RESPONSE -->
.................................................
......!!!!!!!!...................................
18:23:05 gen phy what must happen in order for the sounds from the two speakers to interfere destructively, assuming that the sources are in phase?
......!!!!!!!!...................................
RESPONSE -->
.................................................
......!!!!!!!!...................................
18:23:06 gen phy what must happen in order for the sounds from the two speakers to interfere destructively, assuming that the sources are in phase?
......!!!!!!!!...................................
RESPONSE -->
.................................................
......!!!!!!!!...................................
18:23:16 ** The path difference has to be and integer number of wavelengths plus a half wavelength. **
......!!!!!!!!...................................
RESPONSE -->
.................................................
......!!!!!!!!...................................
18:23:22 CRAB NEBULA PROBLEM?
......!!!!!!!!...................................
RESPONSE -->
.................................................
......!!!!!!!!...................................
18:23:26 This Query will exit.
......!!!!!!!!...................................
RESPONSE -->
.................................................
߉ؐFGy| assignment #026 026. `Query 15 Physics II 07-20-2008
......!!!!!!!!...................................
19:45:03 Principles of Physics and General College Physics Problem 23.08. How far from a concave mirror of radius 23.0 cm must an object be placed to form an image at infinity?
......!!!!!!!!...................................
RESPONSE -->
.................................................
......!!!!!!!!...................................
19:45:46 Recall that the focal distance of this mirror is the distance at which the reflections of rays parallel to the axis of the mirror will converge, and that the focal distance is half the radius of curvature. In this case the focal distance is therefore 1/2 * 23.0 cm = 11.5 cm. The image will be at infinity if rays emerging from the object are reflected parallel to the mirror. These rays would follow the same path, but in reverse direction, of parallel rays striking the mirror and being reflected to the focal point. So the object would have to be placed at the focal point, 11.5 cm from the mirror.
......!!!!!!!!...................................
RESPONSE -->
.................................................
......!!!!!!!!...................................
19:45:57 query gen phy problem 23.14 radius of curvature of 4.5 x lens held 2.2 cm from tooth
......!!!!!!!!...................................
RESPONSE -->
.................................................
......!!!!!!!!...................................
19:46:12 ** if the lens was convex then its focal length would be negative, equal to half the radius. Thus we would have 1 / 2.2 cm + 1 / image distance = -1 / 1.7 cm. Multiplying by the common denominator 1.7 cm * image distance * 1.7 cm we would get 1.7 cm * image distance + 2.2 cm * 1.7 cm = - 2.2 cm * image distance. Thus -3.9 cm * image distance = - 2.2 cm * 1.7 cm. Solving would give us an image distance of about 1 cm. Since magnification is equal to image distance / object distance the magnitude of the magnification would be less than .5 and we would not have a 4.5 x magnification. We have the two equations 1 / image dist + 1 / obj dist = 1 / focal length and | image dist / obj dist | = magnification = 4.5, so the image distance would have to be either 4.5 * object distance = 4.5 * 2.2 cm = 9.9 cm or -9.9 cm. If image dist is 9.9 cm then we have 1 / 9.9 cm + 1 / 2.2 cm = 1/f. Mult by common denominator to get 2.2 cm * f + 9.9 cm * f = 2.2 cm * 9.9 cm so 12.1 cm * f = 21.8 cm^2 (approx) and f = 1.8 cm. This solution would give us a radius of curvature of 2 * 1.8 cm = 3.6 cm, since the focal distance is half the radius of curvature. This positive focal distance implies a concave lens, and the image distance being greater than the object distance the tooth will be more than the focal distance from the lens. For this solution we can see from a ray diagram that the image will be real and inverted. The positive image distance also implies the real image. The magnification is - image dist / obj dist = (-9.9 cm) / (2.2 cm) = - 4.5, with the negative implying the inverted image whereas we are looking for a +4.5 magnification. There is also a solution for the -9.9 m image distance. We eventually get 2.2 cm * f - 9.9 cm * f = 2.2 cm * (-9.9) cm so -7.7 cm * f = -21.8 cm^2 (approx) and f = 2.9 cm, approx. This solution would give us a radius of curvature of 2 * 2.0 cm = 5.8 cm, since the focal distance is half the radius of curvature. This positive focal distance also implies a concave lens, but this time the object is closer to the lens than the focal length. For this solution we can see from a ray diagram that the image will be virtual and upright. The negative image distance also implies the virtual image. The magnification is - image dist / obj dist = -(-9.9 cm) / (2.2 cm) = + 4.5 as required; note that the positive image distance implies an upright image. **
......!!!!!!!!...................................
RESPONSE -->
.................................................
......!!!!!!!!...................................
19:47:19 **** query univ phy problem 33.38 (34.28 10th edition) 3 mm plate, n = 1.5, in 3 cm separation between 450 nm source and screen. How many wavelengths are there between the source and the screen?
......!!!!!!!!...................................
RESPONSE --> The air has a lower index of refraction of 1 and the glass is 1.4. The thickness of the material divided by the thickness per wavelength is the number of wavelengths in the material. so 1.55*10^7 / (540/1) = 28000 wavelengths in air and 2.5*10^-6 / (540/1.4)= 6.5*10^-9 wavelengths in the glass.
.................................................
......!!!!!!!!...................................
23:47:52 ** The separation consists of 1.55 cm = 1.55 * 10^7 nm of air, index of refraction very close to 1, and 2.5 mm = 2.5 * 10^-6 nm of glass, index of refraction 1.4. The wavelength in the glass is 540 nm / 1.4 = 385 nm, approx.. So there are 1.55 * 10^7 nm / (540 nm/wavelength) = 2.27 * 10^4 wavelengths in the air and 2.5 * 10^-2 m / (385 nm/wavelength) = 6.5 * 10^3 wavelengths in the glass. **
......!!!!!!!!...................................
RESPONSE --> I'm not sure why the second answer is different.
.................................................